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Yuki888 [10]
3 years ago
5

Ben is 12 years older than Ishaan. Ben and Ishaan first met two years ago. Three years ago, Ben was 4 times as

Mathematics
1 answer:
sergey [27]3 years ago
4 0

Answer:

Ishaan is now 6.

Step-by-step explanation:

If Ben is 12 years older, and you have the info that <em>three</em> years ago, he was <em>4 </em>times as old, then all you have to do is divide 12 by 4, giving you 3. You add 3 to that because it's been three years. You have your answer of 6.

Hope this helps! :)

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A line passes through the points (2,-4) and (6,10). What is the equation of the line?
deff fn [24]

Answer:

y = \frac{7}{2} x - 11

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (2, - 4) and (x₂, y₂ ) = (6, 10)

m = \frac{10+4}{6-2} = \frac{14}{4} = \frac{7}{2} , thus

y = \frac{7}{2} x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation

Using (2, - 4) , then

- 4 = 7 + c ⇒ c = - 4 - 7 = - 11

y = \frac{7}{2} x - 11 ← equation of line

7 0
3 years ago
Does anyone know the answer to the pic shown above?
Ganezh [65]

Answer:

-1 is the answer

Step-by-step explanation:

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3 years ago
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Answer this for branlist if get right
dusya [7]

Answer:

The median and IQR

Step-by-step explanation:

4 0
3 years ago
What is a 90 degree counter clockwise rotattion on a graph
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<h2><u>Answer:</u></h2><h2>90 Degree Rotation</h2><h2>When rotating a point 90 degrees counterclockwise about the origin our point A(x,y) becomes A'(-y,x). In other words, switch x and y and make y negative.</h2>

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7 0
4 years ago
Find the smallest positive $n$ such that \begin{align*} n &amp;\equiv 3 \pmod{4}, \\ n &amp;\equiv 2 \pmod{5}, \\ n &amp;\equiv
Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number x such that taking it mod 4, 5, and 7 leaves the desired remainders:

x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6

  • Taken mod 4, the last two terms vanish and we have

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4

so we multiply the first term by 3.

  • Taken mod 5, the first and last terms vanish and we have

x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5

so we multiply the second term by 2.

  • Taken mod 7, the first two terms vanish and we have

x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7

so we multiply the last term by 7.

Now,

x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147

By the CRT, the system of congruences has a general solution

n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}

or all integers 27+140k, k\in\mathbb Z, the least (and positive) of which is 27.

3 0
3 years ago
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